Chapter 1: Problem 8

Solve the following equations for \(x\) : a. \(\cosh (x+\ln 3)=3\) b. \(2 \tanh ^{-1} \frac{x-2}{x-1}=\ln 2\). c. \(\sinh ^{2} x-7 \cosh x+13=0\).

Short Answer

Expert verified

\na. \(x = \ln (\frac{-1 \pm 2 \sqrt{10}}{6}) - \ln 3\)\nb. \(x= \frac{3-\sqrt{3}}{2}\)\nc. \(x= \pm acosh (3), \pm acosh (4)\)

Step by step solution

Solve \(\cosh(x + \ln 3)=3\)

First, realize that \(\cosh x = e^x + e^{-x} / 2\). The equation implies that \((e^x * e^{\ln 3} + e^{-x} / e^{\ln 3}) / 2 = 3\). Simplify this to get \(3 e^{2x} - 2e^x -3=0\). This quadratic equation can be solved using the Quadratic formula: \(e^{2x} = \frac{2 \pm \sqrt{4 + 4 * 3 * 3}}{2*3}\) or \(e^{2x} = \frac{-1 \pm 2\sqrt{10}}{3}\). To find \(x\), apply \(\ln\) function to both sides, resulting in \(x = \ln (\frac{-1 \pm 2 \sqrt{10}}{6}) - \ln 3\).Note that only the positive root is considered since \(e^x\) is always positive.

Solve \(2 \tanh^{-1} \frac{x-2}{x-1} = \ln 2\)

Use the fact that \(\tanh^{-1} x = 0.5 \ln \frac{1+x}{1-x}\). Substitute to get \( \ln \frac{(x-1)^2 + (x-2)^2}{(x-1)^2 - (x-2)^2} = \ln 2\). Removing the logarithm yields \( \frac{(x-1)^2 + (x-2)^2}{(x-1)^2 - (x-2)^2} = 2\). Solving for \(x\) gives \(x= \frac{3\pm\sqrt{3}}{2}\), noting that only \(x= \frac{3-\sqrt{3}}{2}\) makes sense, as the inverse hyperbolic tangent has the domain \(-1<x<1\).

Solve \(\sinh^2 x - 7 \cosh x + 13=0\)

Start by using the identity \(\sinh^2 x = \cosh^2 x - 1\), so \(\cosh^2 x - 7 \cosh x + 14=0\). Solving this quadratic gives \(\cosh x = \frac{7\pm\sqrt{49-4*14}}{2} = \frac{7\pm\sqrt{1}}{2}\). This results in two solutions: \(\cosh^2 x = 3\) and \(\cosh^2 x = 4\). These give \(x= \pm acosh (3)\) and \(x= \pm acosh (4)\).